induction proof of fundamental theorem of arithmetic

Proof. If n=1, it is the empty product of primes, and
if n=2, it is a prime number.

Let then n>2. Make the induction hypothesis that all positive
integers m with 1<m<n are products of prime numbers.
If n is a prime number, the thing is ready. Else, n is a
product of smaller numbers; these are, by the induction hypothesis,
products of prime numbers. The proof is complete.

Proof. The assertion is clear in the case that n is a
prime number, especially when n=2.

Let then n>2 and suppose that the assertion is true for
all positive integers less than n.

If now n is a prime, we are ready. Therefore let it be a
composite number. There is a least nontrivial factor p of n.
This p must be a prime. Put n=p⁢b where b is a
positive integer. By the induction hypothesis, b has a
unique prime factor decomposition. Thus n has a unique prime
decomposition containing the prime factor p.

Now we will show that n cannot have other prime decompositions.
Make the antithesis that n has a different prime decomposition;
let q be the least prime factor in it. Now we have p<q
and n=q⁢c where c∈ℤ+ and c<n
with p∤c. Then

n0:=n-p⁢c={p⁢b-p⁢c=p⁢(b-c)q⁢c-p⁢c=(q-p)⁢c

is a positive integer less than n. Since p∣n0, the
induction hypothesis implies that the prime p is in the prime
decomposition of (q-p)⁢c and thus also at least of q-p
or c. But we know that p∤c, whence p∣q-p.
Thus we would get p∣q-p+p=q. Because both p
and q are primes, it would follow that p=q. This
contradicts the fact that p<q. Consequently, our
antithesis is wrong. Accordingly, n has only one prime
decomposition, and the induction proof is complete.